Engineering Mechanics

(^408) „„„„„ A Textbook of Engineering Mechanics
= 1024++ 576 1536 cos (90°+°5 )
= 1600 – 1536 sin 5° ...[ cos (90Q °+θ =−) sin ]θ
=×=1600 – (1536 0.0872) 38.3 km.p.h. Ans.
Example 19.6. A submarine, travelling on a course of bearing 80 degrees East of North
with a speed of 21 knots, reports the presence of an enemy ship travelling on a course of bearing 135
degrees West of North with a speed of 15 knots with respect to the submarine. Assuming that the
submarine and the enemy ship are at the same place at the time of sighting,
(a) Find the true direction and speed of the enemy ship.
(b) If a warship stationed 150 nautical miles South of the submarine immediately starts at
18 knots to intercept the enemy ship, what bearing should it take? Find the time it will
take to intercept the enemy ship.
Solution. Given : Velocity of submarine = 21 knots (N 80°^ E); Velocity of enemy ship = 15
knots (N 135° W) and velocity of warship = 18 knots
(a) True direction and speed of the enemy ship.
(a) Actual velocity diagram (b) Relative velocity diagram
Fig. 19.8.
First of all , let us draw the actual velocity diagram of the submarine and its relative velocity
with enemy ship as shown in Fig. 19.8 (a). Now draw the relative velocity diagram as shown in Fig.
19.8 (b) and as discussed below :

1. First of all, draw the East, West, North and South lines meeting at X.


2. Since the submarine is travelling on a course of bearing 80 degrees East of North,
   therefore, draw a line XL. at 80° to the North representing the actual direction of the
   submarine.


3. Now cut off XM equal to 21 knots to some suitable scale on the opposite direction of the
   actual motion of the submarine.


4. Now draw a line at an angle of 135° West of North and cut off XR equal to 15 knots to the
   scale to represent the relative velocity of the enemy ship with respect to the submarine.


5. Complete the parallelogram XMRN with XM as one side and XR as diagonal.


6. Now the side XN of the diagonal gives the true direction and speed of the enemy ship. By
   measurement, we find that ∠ θ = 35° and XN = 12.7 knots Ans.